CODEWITHSUNDEEP
c
user1547386
user1547386

Reputation: 91

Casting a char * to a char * [] in C

I'm having a problem casting, so is there a way to cast a type of:

char *result;

to a type of

char *argv[100];

?

If so, how would I do this or is there a safe way to do this?

Upvotes: 2

Views: 247

Answers (4)

JackCColeman
JackCColeman

Reputation: 3807

First, a short explanation:

  char *result;

The variable result is a pointer and when set it will point to a single character. However, as you know, a single character can be the start of a string that ends with the null (\0) character.

In C, a good programmer can use the pointer result to index through a string.

However, the string's length is NOT known until the pointer reaches a null character.

It is possible to define a fixed length string of characters, in this case code:

 char s[100];

Now, the fun begins. s per Kernighan and Ritchie (K&R) is a pointer to a string of characters terminated with a 0.

So, you can code:

 s[0] = 'a';
 *s   = 'a';

 s[1] = 'b';
 *(s+1) = 'b';

These are equivalent statements.

As mentioned in other posts, let's add explicit parens to your argv statement:

char *(argv[100]);

Thus, this is an array of 100 pointers to characters (each of which might or might not be the start of a string of characters).

Upvotes: 0

adityajones
adityajones

Reputation: 601

char *result is a pointer to a char
char *argv[100] is an array of char *, so really it's a char ** (a pointer to pointers)

Keep this in mind: int* arr[8]; // An array of int pointers.
int (*arr)[8]; // A pointer to an array of integers

This being the case, this is probably not what you want to be doing. I suppose the next question is: What were you trying to do? Or why?

Upvotes: 2

John Bode
John Bode

Reputation: 123468

What does result contain, and what do you expect argv to contain after the conversion?

For example, if result points to a list of strings separated by a delimiter, like "foo,bar,bletch,blurga,.., and you want each string to be a separated element in argv, like

argv[0] == "foo"
argv[1] == "bar"
argv[2] == "bletch"
argv[4] == "blurga"

then you could not accomplish this with a simple cast; you'd have to actually scan result and assign individual pointers to argv.

Upvotes: 0

sung
sung

Reputation: 366

char * result is a string

and

char * argv[100] is array of strings.

You cannot convert string into array of strings. However, you can create an array of strings where the first array value is result.

argv[0] = result;

Upvotes: 4

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